exact sequenceExactSequence22002-04-24 01:12:402008-06-09 15:11:45Definitionimage morphism% this is the default PlanetMath preamble. as your knowledge
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\newtheorem{definition}[theorem]{Definition}Let $\A$ be an abelian category. We begin with a preliminary definition.
\begin{definition}
For any morphism $f: A \longrightarrow B$ in $\A$, let $m: X \longrightarrow B$ be the morphism equal to $\ker(\cok(f))$. Then the object $X$ is called the {\em image} of $f$, and denoted $\Im(f)$. The morphism $m$ is called the {\em image morphism} of $f$, and denoted $\im(f)$.
\end{definition}
Note that $\Im(f)$ is not the same as $\im(f)$: the former is an object of $\A$, while the latter is a morphism of $\A$. We note that $f$ factors through $\im(f)$:
$$
\xymatrix{
A \ar[r]^-{e} \ar@/_1pc/[rr]_{f} & \Im(f) \ar[r]^-{m} & B
}
$$
The proof is as follows: by definition of cokernel, $\cok(f) f = 0$; therefore by definition of kernel, the morphism $f$ factors through $\ker(\cok(f)) = \im(f) = m$, and this factor is the morphism $e$ above. Furthermore $m$ is a monomorphism and $e$ is an epimorphism, although we do not prove these facts.
\begin{definition}
A sequence
$$
\xymatrix{
\cdots \ar[r] & A \ar[r]^-f & B \ar[r]^-g & C \ar[r] & \cdots
}
$$
of morphisms in $\A$ is {\em exact} at $B$ if $\ker(g) = \im(f)$.
\end{definition}